SOLUTION: Use: f(x) = -2x^2 -5x +3 State the intervals over which f(x) > 0

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Question 1060525: Use: f(x) = -2x^2 -5x +3
State the intervals over which f(x) > 0
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+-2x%5E2+-5x+%2B+3+
f'(x) = +-4x+-5+
f''(x) = +-4+
Since f''(x) < 0, the function is (everywhere) concave down.
Setting f'(x) = 0: +-4x-5+=+0+
+x+=+-5%2F4+
This is a critical point, it is a maximum or minimum. In this case, since the function is concave down, it is a maximum.
So we know f(x) > 0 at the maximum, and it will be greater than zero for values between the zero crossings of f(x). To find the zero crossings, set f(x) = 0:
++-2x%5E2+-+5x+%2B+3+=+0+
++++2x%5E2+%2B+5x+-+3+=+0+
++%282x-1%29%28x%2B3%29+=+0+
++x=1%2F2+ or +x=-3+
So +f%28x%29+%3E+0+ on the interval +-3+%3C+x+%3C+1%2F2+